Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Textbook Question
Chapter 2, Problem 154GQ

Estimating the radius of a lead atom.

  1. (a) You are given a cube of lead that is 1.000 cm on each side. The density of lead is 11.35 g/ cm3. How many atoms of lead are in the sample?
  2. (b) Atoms are spherical; therefore, the lead atoms in this sample cannot fill all the available space As an approximation, assume that 60% of the space of the cube is filled with spherical lead atoms. Calculate the volume of one lead atom from this information. From the calculated volume (V) and the formula (4/3) πτ3 for the volume of a sphere, estimate the radius (r) of a lead atom.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The number of atoms for lead in given sample of lead cube, with each side of value 1cm needed to be calculated.

Concept introduction:

Conversion formula for mass of a molecule and number moles,

Numberofmoles=MassingramsMolarmass

Equation for number of atoms is,

    Number of moles×6.022×1023atoms=number of atoms

Equation for density from volume and mass is,

    Density=MassVolume

Equation for finding Volume of sphere is,

    Volume=(4/3)πr3

Answer to Problem 154GQ

The number of atoms of lead is 3.3×1022atoms

Explanation of Solution

The side of the lead cube is given that 1cm.

Therefore, the volume of the lead cube is,

    (1cm)3=1cm3

The density of the lead cube is given as 11.35g/cm3.

Equation for mass from volume and density is,

    Density×Volume=Mass

Therefore, the mass of lead cube is,

    11.35g/cm3×1cm3=11.35g

Conversion formula for mass of a molecule and number moles,

Numberofmoles=MassingramsMolarmass

Therefore, the number of lead atoms is,

    Numberofmoles=11.35g207.2g/mol=0.05477mol

Equation for number of atoms is,

    Number of moles×6.022×1023atoms=number of atoms

Therefore, the number of lead atoms in the sample is,

    0.05477×6.022×1023atoms=3.3×1022atoms

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The volume of one lead atom and its radius have to be calculated under given conditions.

Concept introduction:

Conversion formula for mass of a molecule and number moles,

Numberofmoles=MassingramsMolarmass

Equation for number of atoms is,

    Number of moles×6.022×1023atoms=number of atoms

Equation for density from volume and mass is,

    Density=MassVolume

Equation for finding Volume of sphere is,

    Volume=(4/3)πr3

Answer to Problem 154GQ

The volume and radius of one lead atom is 1.8×10-23cm3 and 0.7572×10-23cm respectively.

Explanation of Solution

The volume of the lead cube is found that 1cm3.

If 60% of the cube is filled with 3.3×1022 lead atom spheres, then the volume of one lead atom is,

    60100×1cm3×13.3×1022=1.8×10-23cm3

Equation for finding Volume of sphere is,

    Volume=(4/3)πr3

Therefore, the radius of one lead atom is,

    1.8×10-23cm3=(4/3)×3.14×r3 r3=1.8×10-23cm3(4/3)×3.14=0.4342×10-23cm r=0.7572×10-23cm

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Chapter 2 Solutions

Chemistry & Chemical Reactivity

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